Subject: Re: [xsl] concat all items in a sequence From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 31 Aug 2005 14:03:21 +0100 |
<xsl:value-of select="count(.|/root/node[contains(text, 'def')]/preceding-sibling::node)"/> This complains because one <node> contains two <text> elements, <xsl:value-of select="1+count(/root/node[text[contains(.,'def')]]/preceding-sibling::node)"/> Regarding true()=elem/contains(...) it's also really good using the slash operator :) but I can't see how to squeeze it in here... (why the true()= part david?) You wouldn't need to squeeze it, it would just fit, but the version michael (I think) suggested using text[contains(. is better. However, elem/contains(...) is a sequence of booleans and you want to know if any of them is true. as always if you want to know if any item in a sequence is equal to something you canm se = and it does an implicit quantification over the sequence. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
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Re: [xsl] concat all items in a seq, andrew welch | Thread | RE: [xsl] concat all items in a seq, Michael Kay |
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